91Ó°ÊÓ

Perform these steps. a. Find the Spearman rank correlation coefficient. b. State the hypotheses. c. Find the critical value. Use \(\alpha=0.05\). d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Textbook Ranking After reviewing 7 potential textbooks, an instructor ranked them from 1 to 7 , with 7 being the highest ranking. The instructor selected one of his previous students and had the student rank the potential textbooks. The rankings are shown. At \(\alpha=0.05\), is there a relationship between the rankings? $$ \begin{array}{l|ccccccc} \text { Textbook } & \mathrm{A} & \mathrm{B} & \mathrm{C} & \mathrm{D} & \mathrm{E} & \mathrm{F} & \mathrm{G} \\ \hline \text { Instructor } & 1 & 4 & 6 & 7 & 5 & 2 & 3 \\ \hline \text { Student } & 2 & 6 & 7 & 5 & 4 & 3 & 1 \end{array} $$

Step by step solution

01

Assign Ranks

Both the instructor and the student have already provided the ranks for the textbooks. The data is as follows: | Textbook | Instructor Rank | Student Rank | | :-------: | :-------------: | :----------: | | A | 1 | 2 | | B | 4 | 6 | | C | 6 | 7 | | D | 7 | 5 | | E | 5 | 4 | | F | 2 | 3 | | G | 3 | 1 |

02

Calculate the Differences and Square

For each textbook, calculate the difference between the instructor's and the student's ranks. Then square those differences: | Textbook | Rank Diff (d) | (d)^2 | | :------: | :-----------: | :---: | | A | -1 | 1 | | B | -2 | 4 | | C | -1 | 1 | | D | 2 | 4 | | E | 1 | 1 | | F | -1 | 1 | | G | 2 | 4 | Sum of (d)^2 = 16.

03

Compute the Spearman Rank Correlation Coefficient

Use the formula for Spearman's rank correlation coefficient:\[ r_s = 1 - \frac{6 \sum (d^2)}{n(n^2-1)}\]Substitute the values: - \( n = 7 \) (the number of textbooks)- \( \sum (d^2) = 16 \)\[ r_s = 1 - \frac{6 \times 16}{7(49 - 1)} = 1 - \frac{96}{336} = 1 - 0.2857 \approx 0.7143\]

04

State the Hypotheses

The hypotheses for the Spearman rank correlation test are:- Null Hypothesis \( H_0 \): There is no correlation between the instructor's and the student's rankings.- Alternative Hypothesis \( H_a \): There is a correlation between the instructor's and the student's rankings.

05

Find the Critical Value

For a Spearman rank correlation coefficient test at \( \alpha = 0.05 \) with \( n = 7 \), consult a Spearman rank correlation table. The critical value for a two-tailed test is approximately ±0.643.

06

Make the Decision

Compare the computed Spearman rank correlation coefficient \( r_s \approx 0.7143 \) to the critical values ±0.643. Since \( 0.7143 > 0.643 \), we reject the null hypothesis in favor of the alternative hypothesis.

07

Summarize the Results

At the \( \alpha = 0.05 \) significance level, there is enough evidence to conclude that there is a significant correlation between the rankings given by the instructor and the student. This suggests a relationship between how the two individuals ranked the textbooks.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hypothesis Testing

Hypothesis testing is a statistical method that helps us decide between two competing statements, known as hypotheses. In the context of our Spearman rank correlation, we aim to determine if there's a relationship between the instructor's and the student's textbook rankings. We form two hypotheses:

• Null Hypothesis (\( H_0 \)): Assumes there is no correlation between the rankings. In other words, any observed relationship is purely due to chance.
• Alternative Hypothesis (\( H_a \)): Claims that there is a correlation between the rankings.

By performing the test, we can evaluate whether the evidence is strong enough to reject the null hypothesis in favor of the alternative.

Critical Value

The critical value is a key component in hypothesis testing. It sets the threshold for deciding whether to reject the null hypothesis. To find the critical value, one can consult a statistical table relevant to the technique being used — in this case, such as a Spearman rank correlation table. Given the significance level \( \alpha = 0.05 \) and sample size \( n = 7 \), our critical value is \( \pm0.643 \).

This value helps us establish the cut-off points, or boundaries, within which the correlation needs to lie in order to not reject the null hypothesis. If the computed correlation is outside those boundaries, it signals a significant relationship that couldn't have arisen by chance.

Correlation Coefficient

The Spearman rank correlation coefficient is a measure of how well the relationship between two ranked variables can be described by a monotonic function. It's calculated using:

\[ r_s = 1 - \frac{6 \sum (d^2)}{n(n^2-1)} \]

Where \( d \) represents the difference between paired ranks, and \( n \) is the number of observations. In our example:

• \( n = 7 \)
• \( \sum (d^2) = 16 \)
• This gives us \( r_s \approx 0.7143 \)

A correlation coefficient of \( 0.7143 \) indicates a strong positive relationship between the instructor and student rankings—suggesting they ranked the textbooks in a somewhat similar order.

Rank Differences

Rank differences play a critical role in calculating the Spearman rank correlation coefficient. For each subject (in this case, each textbook), calculate the difference between the instructor's rank and the student's rank. These differences (\( d \)) are then squared to get \( d^2 \), helping us eliminate negative values and emphasize larger differences.

Summing up all \( d^2 \) gives us:

• The sum of these differences square, \( \sum (d^2) = 16 \), serves as a vital part in the calculation of the correlation coefficient.
• Smaller sums of \( d^2 \) indicate more similar rankings between the two sets of ranks.

These values help us compute the Spearman rank correlation and assess the degree of agreement between the two rankings.